{wiki=Midpoint_method}

The implicit midpoint rule advances an autonomous ordinary differential equation by
$$
y_{n+1}=y_n+h f\left(\frac{y_n+y_{n+1}}2\right).
$$
It is a second-order, <A-stable>, <time-symmetric numerical method>.

Implicit midpoint rule

For the <Strang splitting>
$$
F(t)=e^{tA/2}e^{tB}e^{tA/2},
$$
reversing $t$ reverses all three factors, so
$$
F(-t)=e^{-tA/2}e^{-tB}e^{-tA/2}=F(t)^{-1}.
$$
It is therefore a <time-symmetric numerical method>. Multiplication of the three <matrix exponential>[exponential] series shows agreement with $e^{t(A+B)}$ through degree two. Equivalently, the <Baker--Campbell--Hausdorff formula> gives an odd modified generator
$$
\log F(t)=t(A+B)+t^3D+O(t^5)
$$
for a matrix $D$ made from nested <commutator>[commutators]. Exponentiating gives
$$
\boxed{F(t)=e^{t(A+B)}+Ct^3+O(t^4)}
$$
for a matrix $C$ depending on $A$ and $B$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 341 / Section A / 5 / b / Solution

Let $\Phi_h$ be the one-step map of a numerical method. It is a <time-symmetric numerical method> when reversing the step exactly reverses the update:
$$
\boxed{\Phi_{-h}=\Phi_h^{-1},
\quad\text{equivalently}\quad
\Phi_{-h}\circ\Phi_h=I.}
$$

Let $\varphi_h$ be the exact <flow map> and suppose the method has order $p$ with a nonzero leading <local truncation error>:
$$
\Phi_h(y)=\varphi_h(y)+h^{p+1}d(y)+O(h^{p+2}).
$$
Inverting this expansion changes the sign of its leading perturbation, so
$$
\Phi_h^{-1}(y)=\varphi_{-h}(y)-h^{p+1}\widetilde d(y)+O(h^{p+2}),
$$
where transport by the exact flow only changes $d$ by $O(h)$ and hence does not affect the leading parity. On the other hand, replacing $h$ by $-h$ in the first expansion gives
$$
\Phi_{-h}(y)=\varphi_{-h}(y)+(-1)^{p+1}h^{p+1}\widetilde d(y)+O(h^{p+2}).
$$
Time symmetry equates these expressions, so $(-1)^{p+1}=-1$. Hence $p+1$ is odd and
$$
\boxed{p\text{ is even}.}
$$

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 341 / 5 / a / Solution

A one-step numerical method with step map $\Phi_h$ is time symmetric when $\Phi_{-h}=\Phi_h^{-1}$. Its leading local-error power is odd, so every nontrivial consistent time-symmetric method has even order.

= Symmetric numerical method
{synonym}


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